What mass of steam initially at 125 degree celcius is needed to warm 250g of water in 100g beaker from 22 degrees celcius to 50 degrees celcius?
Heat loss from M grams steam cooling and condensing to 50 C liquid = heat gained by 250 g water increasing from 22 to 50 C PLUS heat gained by beacker. Write that as an equation and solve for M.
You will need to know the specific heat of steam in the gas phase (about 0.4 cal/g C), as well as the heat of condensation (540 cal/g). You will also need the specific heat of the glass beaker. I will leave the latter up to you to find out.
M*[25*0.4 + 540 + 50*1.0] = (heat gained by original water and beaker)
Solve for M
To find the mass of steam needed to warm the water in the beaker, we can use the equation:
Q = mcΔT
where:
Q is the heat energy transferred
m is the mass of the substance
c is the specific heat capacity of the substance
ΔT is the change in temperature
Let's break down the calculation step by step:
Step 1: Calculate the heat energy needed to warm the water from 22°C to 50°C.
Qwater = mwater * cwater * ΔTwater
The specific heat capacity of water is approximately 4.18 Joules/gram°C.
mwater = 250g (mass of water in the beaker)
cwater = 4.18 J/g°C (specific heat capacity of water)
ΔTwater = 50°C - 22°C = 28°C
Qwater = 250g * 4.18 J/g°C * 28°C
Step 2: Calculate the heat energy provided by the steam.
For steam, we need to consider two things: the heat energy required to raise the steam temperature from 125°C to 100°C and the heat energy required to condense the steam at 100°C.
Qsteam = (msteam * csteam * ΔTsteam) + (msteam * L)
The specific heat capacity of steam (csteam) is approximately 2.03 J/g°C.
The latent heat of vaporization (L) of water is approximately 2260 J/g at 100°C.
ΔTsteam = 100°C - 125°C = -25°C (negative because the temperature is decreasing)
We need to find the mass of steam (msteam), so let's assume it as 'x'.
Qsteam = (x * 2.03 J/g°C * -25°C) + (x * 2260 J/g)
Step 3: Equate the heat energy provided by the steam to the heat energy needed by the water.
Qwater = Qsteam
250g * 4.18 J/g°C * 28°C = (x * 2.03 J/g°C * -25°C) + (x * 2260 J/g)
Solve the equation for 'x', which represents the mass of steam needed.
Note: Steps 1 and 2 are calculated separately to consider the different stages of heat transfer. Step 3 equates the two forms of energy, ensuring that the amount of heat energy gained by the water is equal to the amount of heat energy lost by the steam.